The first one is exothermic because energy is a reactant, the second formula is endothermic
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
between two oppositely charged ions
Explanation:
Haemoglobin consists of heme unit which is comprised of an <u>
</u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.
The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>
The organic ring component of heme is - <u>Porphyrin</u>
Under normal conditions, the central atom of heme is - <u></u>
In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.
Answer:
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